Upper bound of $a_{n+1}=a_n + \frac{1}{a_n}$ [duplicate]
Solution 1:
We have $\displaystyle a_{n+1}^2=a_n^2+\frac{1}{a_n^2}+2$. Hence for $n\geq 1$ $$a_n^2=a_0^2+2n+\sum_{k=0}^{n-1}\frac{1}{a_k^2}$$ This imply that $\displaystyle a_k^2\geq a_0^2+2k$ for $k\geq 1$. Hence $$a_n^2\leq a_0^2+2n+\frac{1}{a_0^2}+\sum_{k=1}^{n-1}\frac{1}{a_0^2+2k}$$
Now $$\sum_{k=1}^{n-1}\frac{1}{a_0^2+2k}\leq \int_0^{n}\frac{dt}{a_0^2+2t}=\frac{1}{2}\log (\frac{2}{a_0^2}n+1)$$ and we are done.